Problem: You have found the following ages (in years) of 5 zebras. The zebras are randomly selected from the 46 zebras at your local zoo: $ 9,\enspace 9,\enspace 31,\enspace 29,\enspace 15$ Based on your sample, what is the average age of the zebras? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 46 zebras, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{92.16} + {92.16} + {153.76} + {108.16} + {12.96}} {{5 - 1}} $ $ {s^2} = \dfrac{{459.2}}{{4}} = {114.8\text{ years}^2} $ We can estimate that the average zebra at the zoo is 18.6 years old. There is a variance of 114.8 years $^2$.